1985 AIME Problems/Problem 2: Difference between revisions
| Line 10: | Line 10: | ||
Thus <math>a^2b=2400, ab^2=5760</math>. Multiplying gets | Thus <math>a^2b=2400, ab^2=5760</math>. Multiplying gets | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
(a^2b)(ab^2)=2400\cdot5760 \\ | (a^2b)(ab^2)&=2400\cdot5760 \\ | ||
a^3b^3=(2^5\cdot3\cdot5^2)(2^7\cdot3^2\cdot5) \\ | a^3b^3&=(2^5\cdot3\cdot5^2)(2^7\cdot3^2\cdot5) \\ | ||
ab=\sqrt[3]{2^{12}\cdot3^3\cdot5^3}=240 \\ | ab&=\sqrt[3]{2^{12}\cdot3^3\cdot5^3}=240 \\ | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Adding gets | Adding gets | ||
Revision as of 12:43, 21 August 2019
Problem
When a right triangle is rotated about one leg, the volume of the cone produced is
. When the triangle is rotated about the other leg, the volume of the cone produced is
. What is the length (in cm) of the hypotenuse of the triangle?
Solution
Let one leg of the triangle have length
and let the other leg have length
. When we rotate around the leg of length
, the result is a cone of height
and radius
, and so of volume
. Likewise, when we rotate around the leg of length
we get a cone of height
and radius
and so of volume
. If we divide this equation by the previous one, we get
, so
. Then
so
and
so
. Then by the Pythagorean Theorem, the hypotenuse has length
.
Solution 2
Let
,
be the
legs, we have the
equations
Thus
. Multiplying gets
Adding gets
Let
be the hypotenuse then
$$ (Error compiling LaTeX. Unknown error_msg)h=
See also
| 1985 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||